pandas 上一组最小/最大值

Pandas previous group min/max( pandas 上一组最小/最大值)

问题描述

在Pandas中，我有如下数据集：

                     Value
2005-08-03 23:15:00   10.5
2005-08-03 23:30:00   10.0
2005-08-03 23:45:00   10.0
2005-08-04 00:00:00   10.5
2005-08-04 00:15:00   10.5
2005-08-04 00:30:00   11.0
2005-08-04 00:45:00   10.5
2005-08-04 01:00:00   11.0
...
2005-08-04 23:15:00   14.0
2005-08-04 23:30:00   13.5
2005-08-04 23:45:00   13.0
2005-08-05 00:00:00   13.5
2005-08-05 00:15:00   14.0
2005-08-05 00:30:00   14.0
2005-08-05 00:45:00   14.5

首先，我希望按日期对数据进行分组，并将每个组的最大值存储在新列中，为此我使用了以下代码：

df['ValueMaxInGroup'] = df.groupby(pd.TimeGrouper('D'))['Value'].transform(max)

现在我想创建另一个列来存储以前的组最大值，因此所需的数据框将如下所示：

                     Value  ValueMaxInGroup  ValueMaxInPrevGroup
2005-08-03 23:15:00   10.5             10.5                  NaN
2005-08-03 23:30:00   10.0             10.5                  NaN
2005-08-03 23:45:00   10.0             10.5                  NaN
2005-08-04 00:00:00   10.5             14.0                 10.5
2005-08-04 00:15:00   10.5             14.0                 10.5
2005-08-04 00:30:00   11.0             14.0                 10.5
2005-08-04 00:45:00   10.5             14.0                 10.5
2005-08-04 01:00:00   11.0             14.0                 10.5
...
2005-08-04 23:15:00   14.0             14.0                 10.5
2005-08-04 23:30:00   13.5             14.0                 10.5
2005-08-04 23:45:00   13.0             14.0                 10.5
2005-08-05 00:00:00   13.5             14.5                 14.0
2005-08-05 00:15:00   14.0             14.5                 14.0
2005-08-05 00:30:00   14.0             14.5                 14.0
2005-08-05 00:45:00   14.5             14.5                 14.0

因此，为了简单地获取前一行的值，我使用了

df['ValueInPrevRow'] = df.shift(1)['Value']

有没有办法获得另一组的最小/最大/f(X)？我假设

df['ValueMaxInPrevGroup'] = df.groupby(pd.TimeGrouper('D')).shift(1)['Value'].transform(max)

但是没有起作用。

推荐答案

使用groupby/agg、shift和merge可以得到想要的结果：

import numpy as np
import pandas as pd
df = pd.DataFrame({'Value': [10.5, 10.0, 10.0, 10.5, 10.5, 11.0, 10.5, 11.0, 14.0, 13.5, 13.0, 13.5, 14.0, 14.0, 14.5]}, index=['2005-08-03 23:15:00', '2005-08-03 23:30:00', '2005-08-03 23:45:00', '2005-08-04 00:00:00', '2005-08-04 00:15:00', '2005-08-04 00:30:00', '2005-08-04 00:45:00', '2005-08-04 01:00:00', '2005-08-04 23:15:00', '2005-08-04 23:30:00', '2005-08-04 23:45:00', '2005-08-05 00:00:00', '2005-08-05 00:15:00', '2005-08-05 00:30:00', '2005-08-05 00:45:00']) 
df.index = pd.DatetimeIndex(df.index)

# This is equivalent to
# df['group'] = pd.to_datetime(df.index.date)
# when freq='D', but the version below works with any freq string, not just `'D'`.
grouped = df.groupby(pd.TimeGrouper('D'))
labels, uniqs, ngroups = grouped.grouper.group_info
df['group'] = grouped.grouper.binlabels[labels]

result = grouped[['Value']].agg(max)
result = result.rename(columns={'Value':'Max'})
result['PreviouMax'] = result['Max'].shift(1)

df = pd.merge(df, result, left_on=['group'], right_index=True)
print(df)

收益率

                     Value      group   Max  PreviouMax
2005-08-03 23:15:00   10.5 2005-08-03  10.5         NaN
2005-08-03 23:30:00   10.0 2005-08-03  10.5         NaN
2005-08-03 23:45:00   10.0 2005-08-03  10.5         NaN
2005-08-04 00:00:00   10.5 2005-08-04  14.0        10.5
2005-08-04 00:15:00   10.5 2005-08-04  14.0        10.5
2005-08-04 00:30:00   11.0 2005-08-04  14.0        10.5
2005-08-04 00:45:00   10.5 2005-08-04  14.0        10.5
2005-08-04 01:00:00   11.0 2005-08-04  14.0        10.5
2005-08-04 23:15:00   14.0 2005-08-04  14.0        10.5
2005-08-04 23:30:00   13.5 2005-08-04  14.0        10.5
2005-08-04 23:45:00   13.0 2005-08-04  14.0        10.5
2005-08-05 00:00:00   13.5 2005-08-05  14.5        14.0
2005-08-05 00:15:00   14.0 2005-08-05  14.5        14.0
2005-08-05 00:30:00   14.0 2005-08-05  14.5        14.0
2005-08-05 00:45:00   14.5 2005-08-05  14.5        14.0

---

这里的主要思想是用groupby/agg代替groupby/transform，这样我们就可以获得

result = grouped[['Value']].agg(max)
result = result.rename(columns={'Value':'Max'})
result['PreviouMax'] = result['Max'].shift(1)
#              Max  PreviouMax
# group                       
# 2005-08-03  10.5         NaN
# 2005-08-04  14.0        10.5
# 2005-08-05  14.5        14.0

然后，所需的DataFrame可以表示为与合并的结果 result在group日期。

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